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Problem solution · Python

Count the Number of Incremovable Subarrays II

Count the Number of Incremovable Subarrays II: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
31 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Count the Number of Incremovable Subarrays II, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 31 lines of Python from the credited upstream file 2972.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCount the Number of Incremovable Subarrays II · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  # Same as 2970. Count the Number of Incremovable Subarrays I  def incremovableSubarrayCount(self, nums: list[int]) -> int:    n = len(nums)    startIndex = self._getStartIndexOfSuffix(nums)    # If the complete array is strictly increasing, the total number of ways we    # can remove elements equals the total number of possible subarrays.    if startIndex == 0:      return n * (n + 1) // 2     # The valid removals starting from nums[0] include nums[0..startIndex - 1],    # nums[0..startIndex], ..., nums[0..n).    ans = n - startIndex + 1     # Enumerate each prefix subarray that is strictly increasing.    for i in range(startIndex):      if i > 0 and nums[i] <= nums[i - 1]:        break      # Since nums[0..i] is strictly increasing, find the first index j in      # nums[startIndex..n) such that nums[j] > nums[i]. The valid removals      # will then be nums[i + 1..j - 1], nums[i + 1..j], ..., nums[i + 1..n).      ans += n - bisect.bisect_right(nums, nums[i], startIndex) + 1     return ans   def _getStartIndexOfSuffix(self, nums: list[int]) -> int:    for i in range(len(nums) - 2, -1, -1):      if nums[i] >= nums[i + 1]:        return i + 1    return 0

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